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1. Solve the following equations:

$(a) 2y + \frac{5}{2} = \frac{37}{2}$

(b) 5t + 28 = 10

(c) a /5 + 3 = 2

(d) q/ 4 + 7 = 5

e) 5 x/2 = –5

(f) $\frac{5}{2}x = \frac{25}{4}$

(g) 7m + 19/2 = 13

(h) 6z + 10 = –2

(i) $\frac{3l}{2} = 2/3$

(j) $\frac{2b}{3}- 5 = 3$

Answers (1)

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$(a) 2y + \frac{5}{2} = \frac{37}{2}$

Transposing $\frac{5}{2}$ to the RHS :

$2y\ =\ \frac{37}{2}\ -\ \frac{5}{2}\ =\ 16$

$y\ =\ 8$

(b) 5t + 28 = 10

Transposing 28 to the RHS and then dividing both sides by 5, we get :

$5t\ =\ 10\ -\ 28\ =\ -\ 18$

$t\ =\ -\frac{18}{5}$

(c) a /5 + 3 = 2

Transposing 3 to the RHS and multiplying both sides by 5, we get :

$\frac{a}{5}\ +\ 3\ =\ 2$

$\frac{a}{5}\ =\ -\ 1$

$a\ =\ -\ 5$

(d) q/ 4 + 7 = 5

Transposing 7 to the RHS and multiplying both sides by 4:

$\frac{q}{4}\ +\ 7\ =\ 5$

$\frac{q}{4}\ =\ -\ 2$

$q\ =\ -\ 8$

(e) 5 x/2 = – 5

Multiplying both sides by $\frac{2}{5}$ :

$x\ =\ -5\times \frac{2}{5}\ =\ -\ 2$

(f) $\frac{5}{2}x = \frac{25}{4}$

Multiplying both sides by $\frac{2}{5}$ :

$x\ =\ \frac{25}{4}\times \frac{2}{5}$

$x\ =\ \frac{5}{2}$

(g) 7m + 19/2 = 13

Transposing $\frac{19}{2}$ to the RHS and then dividing both sides by 7 :

$7m\ =\ 13\ -\ \frac{19}{2}\ =\ \frac{7}{2}$

$m\ =\ \frac{1}{2}$

(h) 6z + 10 = –2

Transposing 10 to the RHS and then dividing both sides by 6, we get :

$6z\ =\ -\ 2\ -\ 10\ =\ -\ 12$

$z\ =\ -\ 2$

(i) $\frac{3l}{2} = 2/3$

Multiplying both sides by $\frac{2}{3}$ ,

$l\ =\ \frac{2}{3}\times \frac{2}{3}\ =\ \frac{4}{9}$

(j) $\frac{2b}{3}- 5 = 3$

Transposing 5 to the RHS and then multiplying both sides by $\frac{3}{2}$

$\frac{2b}{3}\ =\ 8$

$b\ =\ 8\times \frac{3}{2}\ =\ 12$

Posted by

Sanket Gandhi

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